Answer:
[tex]Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})[/tex]
Explanation:
According to the first thermodynamic law, the energy must be conserved so:
[tex]dQ = dU - dW[/tex]
Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.
This equation can be solved by integration between an initial and a final state:
(1) [tex]\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW[/tex]
As per work definition:
[tex]dW = F*dr[/tex]
For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:
[tex]dW = PA*dx[/tex]
Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:
[tex]dW = - P*dV[/tex]
So the third integral in equation (1) is:
[tex]\int\limits^1_2 {- P} \, dV[/tex]
Considering the gas as ideal, the pressure can be calculated as [tex]P = \frac{n*R*T}{V}[/tex], so:
[tex]\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV[/tex]
In this particular case as the systems is closed and the temperature constant, n, R and T are constants:
[tex]\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV[/tex]
Replacion this and solving equation (1) between state 1 and 2:
[tex]\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV[/tex]
[tex]Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})[/tex]
[tex]Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}[/tex]
The internal energy depends only on the temperature of the gas, so there is no internal energy change [tex]U_{2} - U_{1} = 0[/tex], so the heat exchanged to the system equals the work done by the system:
[tex]Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})[/tex]
